3.13.23 \(\int \frac {1}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=14 \[ -\frac {1}{3 b (a+b x)^3} \]

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Rubi [A]  time = 0.00, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {27, 32} \begin {gather*} -\frac {1}{3 b (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(-2),x]

[Out]

-1/(3*b*(a + b*x)^3)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin {align*} \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {1}{(a+b x)^4} \, dx\\ &=-\frac {1}{3 b (a+b x)^3}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 1.00 \begin {gather*} -\frac {1}{3 b (a+b x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(-2),x]

[Out]

-1/3*1/(b*(a + b*x)^3)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a^2 + 2*a*b*x + b^2*x^2)^(-2),x]

[Out]

IntegrateAlgebraic[(a^2 + 2*a*b*x + b^2*x^2)^(-2), x]

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fricas [B]  time = 0.39, size = 35, normalized size = 2.50 \begin {gather*} -\frac {1}{3 \, {\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

-1/3/(b^4*x^3 + 3*a*b^3*x^2 + 3*a^2*b^2*x + a^3*b)

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giac [A]  time = 0.17, size = 12, normalized size = 0.86 \begin {gather*} -\frac {1}{3 \, {\left (b x + a\right )}^{3} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

-1/3/((b*x + a)^3*b)

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maple [A]  time = 0.04, size = 13, normalized size = 0.93 \begin {gather*} -\frac {1}{3 \left (b x +a \right )^{3} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b^2*x^2+2*a*b*x+a^2)^2,x)

[Out]

-1/3/b/(b*x+a)^3

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maxima [B]  time = 1.34, size = 35, normalized size = 2.50 \begin {gather*} -\frac {1}{3 \, {\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")

[Out]

-1/3/(b^4*x^3 + 3*a*b^3*x^2 + 3*a^2*b^2*x + a^3*b)

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mupad [B]  time = 0.03, size = 37, normalized size = 2.64 \begin {gather*} -\frac {1}{3\,a^3\,b+9\,a^2\,b^2\,x+9\,a\,b^3\,x^2+3\,b^4\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^2 + b^2*x^2 + 2*a*b*x)^2,x)

[Out]

-1/(3*a^3*b + 3*b^4*x^3 + 9*a^2*b^2*x + 9*a*b^3*x^2)

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sympy [B]  time = 0.24, size = 37, normalized size = 2.64 \begin {gather*} - \frac {1}{3 a^{3} b + 9 a^{2} b^{2} x + 9 a b^{3} x^{2} + 3 b^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

-1/(3*a**3*b + 9*a**2*b**2*x + 9*a*b**3*x**2 + 3*b**4*x**3)

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